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### Hwk7Solutions

Course: MATH 4320, Fall 2008
School: Georgia Tech
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Word Count: 740

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4320 Math A. D. Andrew Fall 2001 Solutions to Homework Assignment 7 Page 1 We will refer to these two figures. Figure 1 Figure 2 Page 208 Problem 2. I = 0 dx (x 2 + 1) 2 . Since this integral is convergent, we may compute 2I (the principal value) by integrating over the contour shown in Figure 1, and taking the limit as R tends to infinity and then divide this result by 2. Now f (z) = 1 (z 2 + 1)...

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4320 Math A. D. Andrew Fall 2001 Solutions to Homework Assignment 7 Page 1 We will refer to these two figures. Figure 1 Figure 2 Page 208 Problem 2. I = 0 dx (x 2 + 1) 2 . Since this integral is convergent, we may compute 2I (the principal value) by integrating over the contour shown in Figure 1, and taking the limit as R tends to infinity and then divide this result by 2. Now f (z) = 1 (z 2 + 1) 2 has poles of order two at i, and only the pole at i is i . Thus 4 = i 2 i - . 4 inside the contour. Furthermore, Re s( f ,i) = - R -R dx (x 2 + 1) 2 + CR dz (z 2 + 1) 2 We estimate the second integral above as CR dz (z 2 + 1) 2 1 (R 2 - 1) 2 (2 R) which tends to zero as R . Taking the limit we see that Math 4320 A. D. Andrew Fall 2001 Solutions to Homework Assignment 7 Page 2 - 0 dx (x 2 + 1) 2 = 1 2 dx (x 2 + 1) 2 i . = i - = 4 2 Page 208 Problem 4. Again the integral is convergent. Thus we will compute the principal value using the contour in Figure 1 as in the example above. This time, the function f (z) = has simple poles inside the + 1 )( z 2 + 4 ) 1 1 contour at i and 2i with residues - and , respectively. Thus 6i 3i (z z2 2 R -R x 2 dx (x 2 + 1)( x 2 + 4) + CR z2 dz (z 2 + 1)(z 2 + 4) 1 1 = . = 2 i - + 6i 3i 3 Since CR z 2 dz (z 2 + 1)(z 2 + 4) x 2 dx R2 4 R - 5R - 4 ( R ) 0 , we see that = 1 2 0 (x 2 + 1)( x + 4) 2 - x 2 dx (x 2 + 1)( x + 4) 2 = 6 Page 214 Problem 4. I = x x sin(x) dx . + 3 2 0 Since the integrand is even, the integral exists if and only if the principal value of x x 2sin(x) dx does. In order to maintain boundedness in the upper half plane, we + 3 - z ei z will use the function f (z) = 2 , integrate it over the contour shown in z + 3 Figure 1, and then take the imaginary part. The only singularity of f inside the e -2 3 contour is a simple pole at 3 i with residue . We use Jordan's Lemma to 2 show that f (z)dz 0 a sR , and deduce that CR Math 4320 A. D. Andrew Fall 2001 Solutions to Homework Assignment 7 Page 3 x eix 1 x eix dx = x 2 + 3 dx = i e- 2 2 x +3 2 1 - e- 2 2 3 0 3 and hence that I = Page 214 Problem 9. . Find the principal value of - x 2 sin(x) dx . + 4x + 5 This is similar to the last problem. For boundedness reasons, we use ei z , integrate over the contour of Figure 1, and take f (z) = 2 z + 4z + 5 imaginary parts. This time we may handle the integral over the semicircular part of the contour using the usual estimate involving the maximum value of the function and the length of the path. Jordan's Lemma is not required. The only 1 singularity inside the contour is a simple pole at - 2 + i with residue - e-1- 2 i . 2 Thus we find that R limR -R x 2 e ix i -1- 2i dx = - e . + 4x + 5 e 2 sin(x) dx = - sin(2) + 4x + 5 e - x 2 Taking imaginary parts we see that PV Page 218 Problem 1. We convert this to an intergal around the unit circle C by means of the substitutions sin() = and obtain 2 1 1 dz z - , d = 2i z iz 5+ 0 d = 4sin( ) 2z C 2 dz 2 = + 5i z - 2 3 Math 4320 A. D. Andrew Fall 2001 since the integrand has a simple pole at Page 226 Problem 2. Solutions to Homework Assignment 7 Page 4 -i -i with resudue . 2 3 To evaluate 0 xa (x 2 + 1) 2 dx, -1 < a < 3 we will use the contour in Figure 2. za The dimple Cr allows us to avoid the singularity at 0 in f (z) = and ( z2 +...

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